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    Can someone help with the following question on geometry? It's as simple as determining lines/gradients but with surds involved:

    Line l passes A(1, 3√3) and B(2 + √3, 3 + 4√3)

    Give the equation of l in form y=mx+c where m and c are surds in their simplest form.


    I've worked out and corroborated with the book that m = (3 + √3) % (1 + √3)

    Using the extended line equation, I get something complicated like

    y - y1 = m(x - x1)

    so y - 3√3 = ((3 + √3) % (1 + √3)) (x - 1)

    Since m is a fraction, should I try to rationalize or something first?

    The book gives the answer as y = √3x + 2√3 if that helps.

    Thanks for any help
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    (Original post by TimeWalker)
    Can someone help with the following question on geometry? It's as simple as determining lines/gradients but with surds involved:

    Line l passes A(1, 3√3) and B(2 + √3, 3 + 4√3)

    Give the equation of l in form y=mx+c where m and c are surds in their simplest form.

    I've worked out and corroborated with the book that m = (3 + √3) % (1 + √3)

    Using the extended line equation, I get something complicated like

    y - y1 = m(x - x1)

    so y - 3√3 = ((3 + √3) % (1 + √3)) (x - 1)

    Since m is a fraction, should I try to rationalize or something first?

    The book gives the answer as y = √3x + 2√3 if that helps.

    Thanks for any help
    Rationalise the denominator for the gradient first, and it'llall turn out nicely.
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    (Original post by kkboyk)
    Rationalise the denominator for the gradient first, and it'llall turn out nicely.
    Give me a moment, would that be multiplying numerator and denominator by 1 - √3 or simply multiplying both sides by 1 + √3 to cancel it?
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    (Original post by TimeWalker)
    Give me a moment, would that be multiplying numerator and denominator by 1 - √3 or simply multiplying both sides by 1 + √3 to cancel it?
    Yes and you would ahve a whole number as your denominator, which you put in your equation for a straight line
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    (Original post by kkboyk)
    Yes and you would ahve a whole number as your denominator, which you put in your equation for a straight line
    Still stuck. The denominator is either root3, 3 or 1, or something.
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    (Original post by TimeWalker)
    Still stuck. The denominator is either root3, 3 or 1, or something.
    The denominator is just the difference of 2 squares so 1^2 - (root 3)^2 = 1-3 = -2
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    (Original post by TimeWalker)
    Give me a moment, would that be multiplying numerator and denominator by 1 - √3 or simply multiplying both sides by 1 + √3 to cancel it?
    You would indeed multiply both the numerator and denominator by 1-\sqrt{3}
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    (Original post by Zacken)
    You would indeed multiply both the numerator and denominator by 1-\sqrt{3}
    Thanks, I thought so, in that case I'm getting
    m = (3 + √3) % (1 + √3) = (3 - 3) % (1 - 3) = a denominator of -2.

    Is this right?

    So for the line equation, using the coordinates in the OP,

    y - y1 = m (x - x1)

    So y - 3√3 = -2(x - 1), but I can't get from there to the final answer in the textbook, which is y = √3x + 2√3...

    Sorry to drag on it's just that I need to get surds down in combination with geometry..
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    (Original post by TimeWalker)
    Thanks, I thought so, in that case I'm getting
    m = (3 + √3) % (1 + √3) = (3 - 3) % (1 - 3) = a denominator of -2.

    Is this right?

    So for the line equation, using the coordinates in the OP,

    y - y1 = m (x - x1)

    So y - 3√3 = -2(x - 1), but I can't get from there to the final answer in the textbook, which is y = √3x + 2√3...

    Sorry to drag on it's just that I need to get surds down in combination with geometry..
    I get different for bit bolded in quote.. The denominator for the gradient is -2 but you need to use the whole of your rationalised gradient when putting it into straight line formula.
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    (Original post by Elzar)
    I get different for bit bolded in quote.. The denominator for the gradient is -2 but you need to use the whole of your rationalised gradient when putting it into straight line formula.
    Thanks. What to you get if you don't mind me asking?

    I'm getting it from numerator*(1 - √3)
    That is (3 + √3) * (1 - √3). 3 *1 = 3, (√3) * (-√3) = -3
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    (Original post by TimeWalker)
    Thanks. What to you get if you don't mind me asking?

    I'm getting it from numerator*(1 - √3)
    That is (3 + √3) * (1 - √3). 3 *1 = 3, (√3) * (-√3) = -3
    The root 3's shouldn't cancel out on the numerator as they're cancelling out on the denominator.

    So you've done two of the terms from the numerator, 3*1 = 3 and (√3)*(-√3) = -3, but what about the other terms?

    If I write out the denominator in full it might help:

    (1 + √3) * (1 - √3) = 1*1 + 1*(-√3) + 1*(√3) + (√3)*(-√3) = 1 - √3 + √3 - 3 = 1 - 3 = -2.
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    I see. And for numerator I get (when expanding) 3 - √3 + √3 - 3
    Which should cancel to -2√3

    So maybe m = (-2√3) % (-2)

    So using the line equation with the gradient: y-y1 = m(x - x1)

    y -3√3 = ((-2√3) % (-2)) (x - 1)

    Cancel denominator on m which has been rationalized to get:
    -2y +6√3 = -2√3x + 2√3

    -2y = -2√3x + 2√3 - 6√3

    Divide by -2 to isolate y:
    y = √3x - 1 + 3√3

    If I take one from the 3√3, I get the answer from the book which is:
    √3x + 2√3

    Hopefully, if you can follow that, that's the right working. It seems to be a matter of rationalizing first, obviously, and getting the operations right.
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    (Original post by TimeWalker)
    I see. And for numerator I get (when expanding) 3 - 3√3 + √3 - 3
    Which should cancel to -2√3

    So maybe m = (-2√3) % (-2)

    So using the line equation with the gradient: y-y1 = m(x - x1)

    y -3√3 = ((-2√3) % (-2)) (x - 1)

    Cancel denominator on m which has been rationalized to get:
    -2y +6√3 = -2√3x + 2√3

    -2y = -2√3x + 2√3 - 6√3

    Divide by -2 to isolate y:
    y = √3x - 1√3 + 3√3

    If I take one from the 3√3, I get the answer from the book which is:
    √3x + 2√3

    Hopefully, if you can follow that, that's the right working. It seems to be a matter of rationalizing first, obviously, and getting the operations right.
    Altered a couple of things in bold which are probably typos but just in case:

    3*(-√3) in the numerator expansion for -3√3
    and 2√3 divided by -2 equals -√3

    You made it more difficult for yourself by not further simplifying the gradient (doesn't matter really - a right answer's a right answer!). (-2√3)/(-2) is correct but do you see that the -2's would have cancelled each other out?
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    (Original post by Elzar)
    Altered a couple of things in bold which are probably typos but just in case:

    3*(-√3) in the numerator expansion for -3√3
    and 2√3 divided by -2 equals -√3

    You made it more difficult for yourself by not further simplifying the gradient (doesn't matter really - a right answer's a right answer!). (-2√3)/(-2) is correct but do you see that the -2's would have cancelled each other out?
    Ah yes the denominator thing makes it much simpler. The expansion was indeed a typo.

    For the last part of the question it wants showing that l meets the x axis at C (-2, 0)

    So I've simply put y = 0 in the l equation as follows:

    0 = √3x + 2√3
    0 = √3(-2) + 2√3
    0 = -2√3 + 2√3, which is self explanatory since x is given (and substituted in) as -2....
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    (Original post by TimeWalker)
    Ah yes the denominator thing makes it much simpler. The expansion was indeed a typo.

    For the last part of the question it wants showing that l meets the x axis at C (-2, 0)

    So I've simply put y = 0 in the l equation as follows:

    0 = √3x + 2√3
    0 = √3(-2) + 2√3
    0 = -2√3 + 2√3, which is self explanatory since x is given (and substituted in) as -2....
    Usually sub just one value in for those show questions (don't sub in x) and solve for the other value. Sub in just y=0, solve for x, hence line l meets x axis at C(-2,0).
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    (Original post by Elzar)
    Usually sub just one value in for those show questions (don't sub in x) and solve for the other value. Sub in just y=0, solve for x, hence line l meets x axis at C(-2,0).
    Ok. Thanks.
 
 
 
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